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Four similar circles (radius-$R$) are packed in a simple square structure. Let's denote the area between these circles as $S$ - red area in attached image..

Now, let's think about a function $f$ that takes a value of $1$ inside $S$ and $0$ outside of $S$. No periodic boundary conditions are assumed. what analytical function describes the Fourier transform of $f$?

Here is a numerical solution (magnitude in log scale) obtained using Python.

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    Fourier transforms apply to real-valued functions. Is your function 1 on the red area and 0 on the blue area, perhaps? And are you looking for the Fourier transform on $S^1 \times S^1$, or on $\Bbb R \times \Bbb R$, where the function is defined to be $0$ outside the unit square? – 2017-02-02
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    Its not clear what you mean by Fourier transform of an area? – 2017-02-02
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    Do you consider "periodic boundary condition"? i.e. is the right border equivalent to the left? Then your solution would most certainly be linked to Bessel functions. – 2017-02-02
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    Thanks. I edited the question as John Hughes and Rajesh Dachiraju suggested. – 2017-02-02
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    @atrash : Are you working in $R^3$? you mean to say cubic, hence this question. – 2017-02-02
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    @RajeshDachiraju yes I am working in R3. – 2017-02-02
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    @RajeshDachiraju These actually cylinders. Maybe it is better to change cubic to square. – 2017-02-02
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    If you're in $R^3$, then "circles" don't pack cubically -- spheres do (or perhaps cylinders). And you still need to answer the question: periodic boundary condition? Or zero outside this square? Without the answer to that, we cannot help you. – 2017-02-02
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    @JohnHughes and @ the_architect No periodic boundary condition. i.e. zero outside this square. – 2017-02-02
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    @atrash : In your numerical solution, What is it that you are showing in the picture? The magnitude? or the phase? – 2017-02-02
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    @RajeshDachiraju The solution shows the magnitude in log scale. – 2017-02-02

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Fraunhofer diffraction of aperture $S$ by means of Fourier Transform:

\begin{align*} F(\mathbf{k}) &= \iint_S f(x,y)e^{i\mathbf{k}\cdot \mathbf{r}} \, dA \\ &= \int_{-R}^{0} e^{ik_x x} dx \left( \int_{-R+\sqrt{-2Rx-x^2}}^{R-\sqrt{-2Rx-x^2}} e^{ik_y y} \,dy \right) \\ &\quad +\int_{0}^{R} e^{ik_x x} dx \left( \int_{-R+\sqrt{2Rx-x^2}}^{R-\sqrt{2Rx-x^2}} e^{ik_y y} dy \right) \\ &= \int_{-R}^{0} 2\sin \left[k_y \left( \sqrt{-2Rx-x^2}-R \right) \right] \frac{e^{ik_x x}}{k_y} \, dx \\ & \quad + \int_{0}^{R} 2\sin \left[k_y \left( \sqrt{2Rx-x^2}-R \right) \right] \frac{e^{ik_x x}}{k_y} \, dx \\ &= 4 \int_{0}^{R} \sin \left[k_y \left( \sqrt{2Rx-x^2}-R \right) \right] \frac{\cos k_x x}{k_y} \, dx \end{align*}

The integral doesn't have close form.

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    $f(x,y) = 1_{x^2+y^2 < R^2} $ ? You should get a Bessel function for $F(k)$ https://en.wikipedia.org/wiki/Fourier_transform#Two-dimensional_functions – 2017-02-02
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    It's not a circle but looks like an astroid. – 2017-02-02
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    @NgChungTak Thank you very much ! – 2017-02-03